17sec^2θ - 15secθ tanθ - 15 = 0
15secθ tanθ = 17sec^2θ - 15
225 sec^2θ tan^2θ = 289sec^4θ - 510sec^2θ + 225
225sec^4θ - 225sec^2θ = 289sec^4θ - 510sec^2θ + 225
64sec^4θ - 285sec^2θ + 225 = 0
That's just a quadratic in sec^2θ, so just solve it and you have your solution candidates.
However, because we squared things, there may be spurious solutions, so you have to check the values in the original equation.
In radians, I get .159,1.000,2.141,2.982
17sec2 θ − 15tanθsecθ − 15 = 0
17/cos^2 Ų - 15(sinŲ/cosŲ)(1/cosŲ) - 15 = 0
times cos^2 Ų
17 - 15sinŲ - 15cos^2 Ų = 0
17 - 15sinŲ - 15(1 - sin^2 Ų) = 0
15sin^2 Ų - 15sinŲ + 2 = 0
sinŲ = (15 ± √105)/30
sinŲ = .158435 or sinŲ = .841565
Ų = 9.1° or 170.9° or Ų = 57.3° or 122.7°
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